Sifting even number with for and append functions

def sieve(numbers):
odd =
for i in numbers:
if (i//2) != 2:
odd.append(i)
return odd

a = [2, 4, 6, 8, 5]
print(sieve(a))

Output:
[2, 6, 8]


I want this function to sift even numbers out and as you can see my if statement is using floor division to divide the value of i to 2 and if it is not equivalent to 2 then it is an odd number. However, the output I get only retains the even numbers. Why is that?

Because dividing a number by 2 isn't how you check if it's odd... Your code checks if the number is equal to 4 (or 5).
– Aran-Fey
Jun 29 at 10:49

Possible duplicate of How to determine if an integer is even or odd
– Aran-Fey
Jun 29 at 10:50

Try it with modulo: if i%2 != 0:
– Igle
Jun 29 at 10:54

if i%2 != 0:

1 Answer
1

I think you have a confusion with // operator. The // is a floor division operator which is very similar to the division operator /. It returns the result of division without floating numbers. You can check here:

//

//

/

>>> a = [2, 4, 6, 8, 5]
>>> for i in a:
print i//2


Output:

1
2
3
4
2


In order to find odd/even numbers you can use modulo operator % as @lgle said:

%

>>> for i in a:
print i%2


Output:

0
0
0
0
1


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